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Question

Let ¯¯bz+b¯¯¯z=c,b0, be a line in the complex plane, where ¯¯b is the complex conjugate of b. If a point z1 is the reflection of a point z2 through the line, then prove that c=¯¯¯z1b+z2¯¯b.

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Solution

The given line is ¯¯bz+b¯¯¯z=c ...(1)
Let A(z1) be a reflection of B(z2) in the line (1)
Let P(z) be any point on the line (1), we have
AP=BP|AP|2=|BP|2|zz1|2=|zz2|2(zz1)(¯¯¯z¯¯¯¯¯z1)=(zz2)(¯¯¯z¯¯¯¯¯z2)
(¯¯¯¯¯z2¯¯¯¯¯z1)z+(z2z1)¯¯¯z+z1¯¯¯zz2¯¯¯¯¯z2=0 ...(2)
Since (1) represent (2) the same line, we get
¯¯b¯¯¯¯¯z2¯¯¯¯¯z1=bz2z1=cz1¯¯¯¯¯z1z2¯¯¯¯¯z2=k(let)
k(¯¯¯¯¯z2¯¯¯¯¯z1)=¯¯b,k(z2z1)=b,k(z1¯¯¯¯¯z1z2¯¯¯¯¯z2)=c
Now ¯¯¯¯¯z1b+z2¯¯b
=¯¯¯¯¯z1[k(z2z1)]+z2(k(¯¯¯¯¯z2¯¯¯¯¯z1))=k(¯¯¯¯¯z1z2z1¯¯¯¯¯z1+z2¯¯¯¯¯z2z2¯¯¯¯¯z1)=k(z2¯¯¯¯¯z2z1¯¯¯¯¯z1)=c

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